1. ## please check my working, trig indentity proof

Can someome please check my working and tell me if it is 'correct'. As my book shows a different method.

using $\displaystyle cos2A = 2cos^{2}A - 1 = 1-2sin^{2}A$

show that; $\displaystyle cos^{2}\frac{x}{2} = \frac{1+cosx}{2}$

my method;

$\displaystyle cosx = cos2\frac{x}{2} = 2cos^{2}\frac{x}{2}-1$

$\displaystyle \frac{1+ 2cos^{2}\frac{x}{2}-1}{2}$ The one's cancel, and I get

$\displaystyle \frac{2cos^{2}\frac{x}{2}}{2}$

and now can I cancel the two's to give me

$\displaystyle cos^{2}\frac{x}{2}$

or do I have to cancel both 'two's at the top?

thanks

2. Originally Posted by Tweety
Can someome please check my working and tell me if it is 'correct'. As my book shows a different method.

using $\displaystyle cos2A = 2cos^{2}A - 1 = 1-2sin^{2}A$

show that; $\displaystyle cos^{2}\frac{x}{2} = \frac{1+cosx}{2}$

my method;

$\displaystyle cosx = cos2\frac{x}{2} = 2cos^{2}\frac{x}{2}-1$

$\displaystyle \frac{1+ 2cos^{2}\frac{x}{2}-1}{2}$ The one's cancel, and I get

$\displaystyle \frac{2cos^{2}\frac{x}{2}}{2}$

and now can I cancel the two's to give me

$\displaystyle cos^{2}\frac{x}{2}$

or do I have to cancel both 'two's at the top?

thanks
You got to $\displaystyle \cos{2\left(\frac{x}{2}\right)} = 2\cos^{2}\frac{x}{2}-1$

$\displaystyle \cos{x} = 2\cos^2{\frac{x}{2}} - 1$

$\displaystyle \cos{x} + 1 = 2\cos^2{\frac{x}{2}}$

$\displaystyle \frac{\cos{x} + 1}{2} = \cos^2{\frac{x}{2}}$.

3. Originally Posted by Prove It
You got to $\displaystyle \cos{2\left(\frac{x}{2}\right)} = 2\cos^{2}\frac{x}{2}-1$

$\displaystyle \cos{x} = 2\cos^2{\frac{x}{2}} - 1$

$\displaystyle \cos{x} + 1 = 2\cos^2{\frac{x}{2}}$

$\displaystyle \frac{\cos{x} + 1}{2} = \cos^2{\frac{x}{2}}$.
Thanks, I understand what you did, however I dont quite understand what I did wrong?

could you please show/point out to me? cause both methods 'looks' valid to me, is it the last part where I cancel 2?

thanks.

4. Hello Tweety
Originally Posted by Tweety
Thanks, I understand what you did, however I dont quite understand what I did wrong?

could you please show/point out to me? cause both methods 'looks' valid to me, is it the last part where I cancel 2?

thanks.
There's nothing wrong with your original method. You cancelled the 2's correctly.